Question

Need help with linear algebra question!! Determine whether linear operator T:R^2 --> R^2 defined by the equation is one to one if so find the standard matrix for the inverse operator and find T^-1(w_1, w_2)

Answer 1

Don't get how to find if it's one to one..

Answer 2

What is T?

Answer 3

Transformation? It's just showing that it's going from R^2 to R^2

Answer 4

its says its defined by the equation....what equation?

Answer 5

HAHA oh snap yeahh

Answer 6

w_1 = 4x_1 - 6x_2 and w_2=2x_1+3x_2

Answer 7

So we have:
\[T(x_1,x_2) = (4x_1-6x_2, 2x_1+3x_2)\]
To show that something is one-to-one, you need to prove that:
\[T(a_1,a_2) = T(b_1,b_2) \Rightarrow (a_1,a_2) = (b_1,b_2)\]

Answer 8

So lets start with the left hand side and see if we can produce the right hand side:
\[T(a_1,a_2) = T(b_1,b_2)\]
Using the definition of the transformation (the formula):
\[T(a_1,a_2) = T(b_1,b_2) \Rightarrow (4a_1-6a_2, 2a_1+3a_2) = (4b_1-6b_2, 2b_1+3b_2)\]

Answer 9

Sorry I can't see the formula so it's T(a_1, a_2) = T(b_1, b_2) --->(4a_1-6a_2, 2a_1+3a_2) = (4b_1-6b_2, 2b_1+3b_2)?

Answer 10

Right. Am i doing something wrong that you cant see the formula?

Answer 11

no it's just a lot of brackets and it says rightarrow haha but it's alright. Okay so how do you know a = b?

Answer 12

Next, lets subtract both sides of that equation by the right hand side:
\[(4a_1-6a_2, 2a_1+3a_2) = (4b_1-6b_2, 2b_1+3b_2) \Rightarrow \]
\[(4a_1-6a_2, 2a_1+3a_2) - (4b_1-6b_2, 2b_1+3b_2) = 0\]
combining them we get:
\[((4a_1-6a_2)-(4b_1-6b_2), (2a_1+3a_2)-(2b_1+3b_2)) = (0, 0)\]

Answer 13

Basically what you want to do is move things around, get a bunch of terms like (a_1 - b_1), (a_2-b_2), etc, and show that because the right hand side is 0, each of those terms must be 0. You end up with each a_i-b_i = 0 which is the same as a_1 = b_i

Answer 14

i'll write the rest out on paper, one sec.

Answer 15

can you just combine the a_1 and b_1 like that?

Answer 16

like what?

Answer 17

Like it was originally this right (4a_1-6a_2, 2a_1+3a_2) - (4b_1-6b_2, 2b_1+3b_2) = 0

Answer 18

then you switch 2a_1 +3a_2 with 4b..etc to get (4a_1-6a_2)-(4b_1-6b_2), (2a_1+3a_2)-(2b_1+3b_2)) = (0, 0)

Answer 19

yes, you could have combined them like that. my professors yell at me if i skip steps so ive gotten accustomed to showing everything lol

Answer 20

Oh okay didn't know you could, no your steps really do help and sorry but do also know what it means when it says T^-1(w_1, w_2)?

Answer 21

scanning a pic, one sec

Answer 22

Oh okay I got up to finding the inverse of A then didnt know what to do.. thanks so much

Answer 23

Hey joemath.. do you think I could trouble you with one more question you seem to know your linear algebra quite well

Answer 24

Go ahead, shoot away :)

Answer 25

okay this has to with basis and dimensions find the canonical basis for the solution space of the homogeneous system and state the dimension of the space..

Answer 26

x_1 + x_2 + x_3 =0; 2x_1-x_2+4x_3 =0; 3x_1 + x_2 + 11x+3 =0

Answer 27

I don't understand dimension of space

Answer 28

Alright, im going to post something in a sec. Basically what you want to do is create the matrix that represents those equations, row reduce it, and look for the pivot and free variables.

Answer 29

Ignore whats in the top left and top right.

Answer 30

Haha okay.. I wrote the equation down wrong but it's alright

Answer 31

so after you get the equations

Answer 32

you put it into a vector form?

Answer 33

yes. make the matrix that would represent that system. Then row reduce it.

Answer 34

oh okay I think I got it thanks!

Answer 35

no prob :)